At 600 K and 1.75 atm, if 16.00 grams of H2 will reacts with excess N2 what volume in liters will be produced of NH3?
N2(g) + 3 H2(g) --> 2 NH3(g)
(R = 0.0821 L atm/mol K)

Answer : The volume of produced will be, 150.0 L

Explanation :

First we have to calculate the moles of

Now we have to calculate the moles of

The balanced chemical equation is:

From the reaction, we conclude that

As, 3 moles of react to give 2 moles of

So, 8 moles of react to give mole of

Now we have to calculate the volume of

Using ideal gas equation:

PV = nRT

where,

P = pressure of gas = 1.75 atm

V = volume of gas = ?

n = number of moles of gas = 5.33 mol

T = temperature of gas = 600 K

R = gas constant = 0.0821 L.atm/mol.K

Now put all the given value in the above formula, we get:

Therefore, the volume of produced will be, 150.0 L